davidwang's Blog

题意： 给你一个数n(n<10^51)，让你把他的各位数字重拍列以下，然后比他小，求方案数

非常简单的数位DP 主要是看一下代码高亮的情况

//coder : davidwang
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <ctime>
#include <vector>
#include <deque>
#include <queue>
using namespace std;
#define LL long long 
#define X first
#define Y second
inline int read()
{
	int ret=0,f=1; char ch;
	for (ch=getchar();ch<'0' || ch>'9';ch=getchar())
		if (ch=='-') f=-f;
	for (;'0'<=ch && ch<='9';ch=getchar())
		ret=ret*10+ch-'0';
	return ret*f;
}
const int N = 60;
LL ans,c[N][N],num[N];
char s[N];
inline LL calc(int x){
	LL ret=1;
	for (int i=0;i<=9;i++)
		ret*=c[x][num[i]],x-=num[i];
	return ret;
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	scanf("%s",s+1);
	int n=strlen(s+1);
	for (int i=1;i<=n;i++) num[s[i]-'0']++;
	c[0][0]=1;
	for (int i=1;i<=n;i++){
		c[i][0]=1;
		for (int j=1;j<=i;j++) c[i][j]=c[i-1][j]+c[i-1][j-1];
	}
	for (int i=1;i<=n;i++){
		for (int j=0;j<s[i]-'0';j++){
			if (!num[j]) continue;
			num[j]--;
			ans+=calc(n-i);
			num[j]++;
		}
		num[s[i]-'0']--;
	}
	printf("%lld\n",ans);
	return 0;
}